Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

namespace {

    typedef double real;
    typedef long long ll;

    int N, M, S, T;
    vector< vector<int> > G;
    void input() {
        cin >> N >> M >> S >> T;
        S--; T--;
        G.resize(N);
        for (int i = 0; i < M; i++) {
            int x, y; cin >> x >> y; x--; y--;
            G[x].push_back(y);
            G[y].push_back(x);
        }
    }

    const int INF = 1<<28;

    void dist(int s, vector<int>& d) {
        d.clear(); d.resize(N, INF);
        d[s] = 0;
        queue< pair<int, int> > Q;
        Q.push(make_pair(s, 0));
        while (not Q.empty()) {
            pair<int, int> c = Q.front(); Q.pop();
            for (int i = 0; i < G[c.first].size(); i++) {
                int n = G[c.first][i];
                if (d[n] > c.second + 1) {
                    d[n] = c.second + 1;
                    Q.push(make_pair(n, c.second + 1));
                }
            }
        }
    }

    void solve() {
        vector<int> d1; dist(S, d1);
        vector<int> d2; dist(T, d2);
        int d = d1[T];
        vector<int> D = d2;
        sort(D.begin(), D.end());
        ll ans = 0;
        for (int i = 0; i < N; i++) {
            int k = d - d1[i] - 2;
            if (k < 0) continue;
            ans += upper_bound(D.begin(), D.end(), k) - lower_bound(D.begin(), D.end(), k);
        }
        cout << ans << endl;
    }
}

int main() {
    input(); solve();
    return 0;
}